Practice question
Question
For 2HI(g) <=> H₂(g) + I₂(g) , Kp = 0.04 at 500 K. If 1 mole of HI is placed in a 1 L vessel, what is the total pressure at equilibrium ( R = 0.0831 bar L/mol K )?
Explanation
Initial: PHI = (1 × 0.0831 × 500/1) = 41.55 bar . Let 2x dissociate, PHI = 41.55 - 2x , PH₂ = PI₂ = x , total pressure = 41.55 - 2x + 2x = 41.55 . Kp = (PH₂ PI₂/(PHI)²) = (x²/(41.55 - 2x)²) = 0.04 , (x/41.55 - 2x) = 0.2 , x ≈ 7.58 bar , total = 41.55 bar.
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